ade7566 Analog Devices, Inc., ade7566 Datasheet - Page 48

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ade7566

Manufacturer Part Number
ade7566
Description
Single-phase Energy Measurement Ic With 8052 Mcu, Rtc, And Lcd Driver
Manufacturer
Analog Devices, Inc.
Datasheet

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ADE7566/ADE7569
PHASE COMPENSATION
The ADE7566/ADE7569 must work with transducers that can
have inherent phase errors. For example, a phase error of 0.1° to
0.3° is not uncommon for a current transformer (CT). These
phase errors can vary from part to part, and they must be corrected
to perform accurate power calculations. The errors associated
with phase mismatch are particularly noticeable at low power
factors. The ADE7566/ADE7569 provide a means of digitally
calibrating these small phase errors. The part allows a small
time delay or time advance to be introduced into the signal
processing chain to compensate for small phase errors. Because
the compensation is in time, this technique should only be used
for small phase errors in the range of 0.1° to 0.5°. Correcting
large phase errors using a time shift technique can introduce
significant phase errors at higher harmonics.
The phase calibration register (PHCAL[7:0]) is a twos complement,
signed, single-byte register that has values ranging from 0x82
(−126d) to 0x68 (+104d).
The PHCAL register is centered at 0x40, meaning that writing
0x40 to the register gives 0 delay. By changing this register, the
time delay in the voltage channel signal path can change from
−231.93 μs to +48.83 μs (MCLK = 4.096 MHz). One LSB is
equivalent to 1.22 μs (4.096 MHz/5) time delay or advance. A
line frequency of 60 Hz gives a phase resolution of 0.026° at the
fundamental (that is, 360° × 1.22 μs × 60 Hz).
Figure 54 illustrates how the phase compensation is used to
remove a 0.1° phase lead in current channel due to the external
transducer. To cancel the lead (0.1°) in current channel, a phase
lead must also be introduced into voltage channel. The resolution
of the phase adjustment allows the introduction of a phase lead
in increments of 0.026°. The phase lead is achieved by introducing
a time advance into the voltage channel. A time advance of 4.88 μs
is made by writing −4 (0x3C) to the time delay block, thus
reducing the amount of time delay by 4.88 μs, or equivalently, a
phase lead of approximately 0.1° at a line frequency of 60 Hz
(0x3C represents −4 because the register is centered with 0 at 0x40).
I
V
I
V
I
V
AP
N
P
N
I
PGA1
PGA2
V
60Hz
ADC 2
0.1°
ADC 1
Figure 54. Phase Calibration
1
–231.93µs TO +48.83µs
7
1
DELAY BLOCK
0
1.22µs/LSB
PHCAL[7:0]
0
1
HPF
0
1
1
24
1
0
CHANNEL 2 DELAY
REDUCED BY 4.48µs
(0.1°LEAD AT 60Hz)
0x0B IN PHCAL[7:0]
24
V
60Hz
I
LPF2
Rev. 0 | Page 48 of 136
RMS CALCULATION
The root mean square (rms) value of a continuous signal V(t) is
defined as
For time sampling signals, rms calculation involves squaring the
signal, taking the average, and obtaining the square root. The
ADE7566/ADE7569 implement this method by serially squaring
the input, averaging them, and then taking the square root of
the average. The averaging part of this signal processing is done
by implementing a low-pass filter (LPF3 in Figure 55, Figure 56,
and Figure 57). This LPF has a −3 dB cut-off frequency of 2 Hz
when MCLK = 4.096 MHz.
where V is the rms voltage.
When this signal goes through LPF3, the cos(2ωt) term is attenu-
ated and only the dc term V
The I
the WAVMODE register (0x0D) and setting the WFSM bit in
the Interrupt Enable 3 SFR (MIRQENH, 0xDB). Like the
current and voltage channels waveform sampling modes, the
waveform date is available at sample rates of 25.6 kSPS,
12.8 kSPS, 6.4 kSPS, or 3.2 kSPS.
It is important to note that when the current input is larger than
40% of full scale, the I
represent the true processed rms value. The rms value processed
with this level of input is larger than the 24-bit read by the wave-
form register, making the value read truncated on the high end.
V
V
V
rms
rms
( )
2
t
signal can be read from the waveform register by setting
) (
V (t ) = √2 × V sin(ωt )
INPUT
t
=
=
=
V
2
T
1
2
×
×
V
T
0
V
Figure 55. RMS Signal Processing
V
sin(
2
2
V
rms
cos
(
2
ω
t
(t) = V
)
waveform sample register does not
t
(
dt
)
2
ω
rms
2
t
– V
)
2
goes through (see Figure 55).
2
cos (2ωt)
LPF3
V
2
(t ) = V
2
V
(1)
(2)
(3)

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